fX,Y(x,y) = 6(1-x-y) for x and y defined over the unit square, subject to the restriction that 0<x+y<1. Find the marginal probability function for X.
I have started by integrating with limits 1-x and -x getting a final answer of 18(x-1-2x^2). Anyone else get this answer?
Thanks
It would help a lot if you show your working. It'd be far easier to spot out your mistake.
I've just made the integration again, and I still found 3(1-x)²... :s
If Mr F didn't say anything, then there is some chance that it is correct
And why do you want to multiply by (6-6x) ? o.O
(6-6x) is not a constant that you simply pull out from the integral... And it doesn't even multiply the integrand !
If a is a constant, then , but that's all you can say !
You can pull out 6, nothing else.
It is true that (1-x) is a constant, but only with respect to y.
You can separate the integral :
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